Solution
= Solution
The <number of e-folds> $N=\log a$ satisfies
$$
dN=d\log a=H\,dt.
$$
When $\phi$ is monotone, the <chain rule> gives $dt=d\phi/\dot\phi$, and hence
$$
dN=\frac{H}{\dot\phi}\,d\phi.
$$
Using the two <slow-roll equations>, $3H\dot\phi\simeq-V'(\phi)$ and $H^2\simeq8\pi GV/3$, eliminates both $H$ and time:
$$
\boxed{dN\simeq-8\pi G\frac{V(\phi)}{V'(\phi)}\,d\phi}.
$$
This is the differential form of the <slow-roll e-fold count>.